$\begin{array}{1 1}(A)\;17153136\\(B)\;170848\\(C)\;1767532\\(D)\;\text{None of these}\end{array} $

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Total no of mice =18

Total no of groups =3

The groups are equally large hence each group can have =6 mice

Hence the total no of ways in which the mice can be placed =18!

Since the possible no of ways the mice can be placed in each group $\Rightarrow 6!$

Hence the total no of ways in which the mice can be placed into three groups are =$\large\frac{18!}{6!\times 6!\times 6!}$

$\Rightarrow \large\frac{18\times 17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 1\times 6\times 5\times 4\times 3\times 2\times 1\times 6!}$

$\Rightarrow 17153136$

Hence (A) is the correct answer.

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